Chemistry

Thermochemistry and Equilibrium

110 Questions

Thermochemistry and equilibrium problems focus on calculating bond energies, lattice enthalpy, and the Born Haber cycle. These concepts are vital for scoring well in chemistry sections. Regular practice ensures a clear understanding of energy changes in reactions.

Bond energy calculationsBorn Haber cycleLattice enthalpyEnthalpy of solutionThermochemical equations

Thermochemistry and Equilibrium Questions

Multiple choice chemistry the world of carbon organic compounds allotropes of carbon versatile nature of carbon

$C _{graphite}+O _{2(g)}\rightarrow CO _{2(g)}; \Delta H=-393.5kJ, \Delta H$ of the reaction cannot be:

  1. Heat of formation of $CO _{2}$
  2. Heat of combustion of $CO _{2}$

  3. Heat of reaction

  4. Heat of transition
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$C(graphite)+O _2(g)\longrightarrow CO _2(g)$


Now, here $\Delta H$ of the reaction cannot be the heat of transition, because there is no transition between phase take place at constant temperature and pressure.

For example: transition of $C _{diamond}\to C _{graphite}$

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

The enthalpies of combustion of carbon and carbon monoxide are -393.5 KJ and -283 KJ respectively the enthalpy of formation of carbon monoxide is :

  1. -676.5 KJ

  2. -110.5 KJ

  3. 110.5 KJ

  4. 676.5 KJ
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The enthalpy of formation of carbon monoxide is calculated using Hess's law by subtracting the enthalpy of combustion of carbon monoxide from that of carbon. Specifically, delta H_f(CO) = delta H_c(C) - delta H_c(CO) = -393.5 - (-283) = -110.5 kJ. This represents the energy change when one mole of carbon monoxide is formed from its elements.

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

The enthalpy of tetramerization of $X$ in gas phase $(4X(g)\rightarrow { X } _{ 4 }(g))$ is $-100\ kJ/mol$ at $300\ K$. The enthalpy of vaporisation for liquid $X$ and ${X} _{4}$ are respectively $30\ kJ/mol$ and $72\ kJ/mol$ respectively.
$\Delta S$ for tetramerization of $X$ in liquid phase is $-125\ J/K mol$ at $300\ K$.
What is the $\Delta G$ at $300\ K$ for tetramerization of $X$ in liquid phase?

  1. $-52\ kJ/mol$
  2. $-98\ kJ/mol$
  3. $-14.5\ kJ/mol$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

Heat of formation of $2$ moles of ${NH} _{3}(g)$ is $-90kJ$; bond energies of $H-H$ and$N-N$ bonds are $435kJ$ and $390kJ$ ${mol}^{-1}$ respectively. The value of the bond energy of $N\equiv N$ will be:

  1. $-472.5\ kJ$
  2. $-945\ kJ$
  3. $472.5\ kJ$
  4. $945\ kJ$ ${mol}^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\Delta { H } _{ reaction }=\sum { { \left( BE \right)  } _{ reactants } } -\sum { { \left( BE \right)  } _{ products } } $
$-90=x+3\times 435-6\times 390$
$x=945kJ$ ${mol}^{-1}$

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

The standard enthalpy of formation of ${NH} _{3}$ is $-46kJ{ mol }^{ -1 }$. If the enthalpy of formation of ${H} _{2}$ from its atoms is $-436kJ{ mol }^{ -1 }$ and that of ${N} _{2}$ is $-712kJ{ mol }^{ -1 }$, the average bond enthalpy of $N-H$ bond in ${NH} _{3}$ is:

  1. $+1056kJ{ mol }^{ -1 }\quad $
  2. $-1102kJ{ mol }^{ -1 }\quad $
  3. $-964kJ{ mol }^{ -1 }\quad $
  4. $+352kJ{ mol }^{ -1 }\quad $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

${ H } _{ reaction }=\sum { { BE } _{ reactants } } -\sum { { BE } _{ products } } $
$-46=(\cfrac(712)+\cfrac{3}{2}(436))-3x$
$x=+352\ kJ{ mol }^{ -1 } $

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

Calculate $ \Delta { H }^{ o }$ of the reaction:

$ { CH } _{ 2 }={ CH } _{ 2 }+3O=O\longrightarrow 2O=C=O+2H-O-H$

The average bond enthelpies of various bond are:

$Bond \quad \quad \quad\quad \quad C-H\quad \quad O=O\quad \quad C=O\quad \quad O-H\quad \quad C=C$
$Bond\ enthalpy$:     414               499               724           460               619
[kJ/mol]

  1. -364kJ

  2. -564kJ

  3. -964kJ

  4. -1654kJ
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using bond enthalpy method: ΔH = Σ(bonds broken) - Σ(bonds formed). Breaking: 4 C-H (4×414=1656), 1 C=C (619), 3 O=O (3×499=1497). Total broken = 3772. Forming: 4 C=O (4×724=2896), 4 O-H (4×460=1840). Total formed = 4736. ΔH = 3772-4736 = -964 kJ. The reaction is exothermic.

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

$H _{2}+Cl _{2}\rightarrow 2HCl+44\ K.Cal$. Heat of decomposition of $HCl$ is:

  1. $-44\ K.Cal$
  2. $+44\ K.Cal$
  3. $-22\ K.Cal$
  4. $+22\ K.Cal$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$H _2+Cl _2 \longrightarrow 2HCl+44Kcal$

Heat of decomposition of $HCl$ :-
$2HCl \longrightarrow \Delta H _{decomp}=+44 Kcal$
$1 HCl \longrightarrow \Delta H _{decomp}=+22 Kcal$
So, Heat of decomposition of $HCl=+22Kcal$

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

The standard enthalpies of n-pentane, isopentane and neopentane are $-35.0,\ -37.0$ and $-40.0$ $K \ cal/mole$ respectively. The most stable isomer of pentane in terms of energy is ____________.

  1. n-pentane

  2. isopentane

  3. neopentane

  4. n-pentane and isopentane
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The standard enthalpies of n-pentane, isopentane and neopentane are -35.0, -37.0 and -40.0 K.cal/mole respectively. The most stable isomer of pentane in terms of energy is neopentane as it has most negative value of the standard enthalpy.

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

Calculate P - CI bond enthalpy 
Given : $\Delta f H(PCl _3, g) = 306 KJ/mol;$     $\Delta H _{atomization} (P, s) = 314 KJ / mol;$
$\Delta f H (Cl, g) = 121 KJ / mol$

  1. 123.66 KJ/mol

  2. 371 KJ / mol

  3. 19 KJ/ mol

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Born-Haber type approach: ΔfH(PCl3) = ΔHatom(P) + 3ΔHatom(Cl) - 3BE(P-Cl). BE(P-Cl) = [314 + 3(121) - 306]/3 = 123.67 kJ/mol. The calculation uses formation enthalpy as the difference between atomization and bond formation energies.

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

Calculate the average Bond energy of O-F bond in the following reaction :-
$OF _{2(g)}\rightarrow O _{(g)}+2F _{(g)}$
Given :
$OF _{2(g)}\rightarrow OF _{(g)}+F _{(g)}$; $\Delta H$=201 kJ
$OF _{(g)}\rightarrow O _{(g)}+F _{(g)}$; $\Delta H$=199 kJ

  1. 201 kJ

  2. 199 kJ

  3. 200 kJ

  4. 200.9 kJ
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The overall reaction for atomization of OF2 into one oxygen and two fluorine atoms is the sum of the two given steps, so its total enthalpy change is 201 + 199 = 400 kJ. Since OF2 contains two O-F bonds, the average bond energy is obtained by dividing the total enthalpy by two, yielding 400 / 2 = 200 kJ.

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

Which one of the following statement(s) is/are true?

  1. $\Delta E=0$ for combustion of ${ C } _{ 2 }{ H } _{ 6 }(g)$ in a sealed rigid adiabatic container
  2. ${ \Delta } _{ f }{ H }^{ o }(S,monolithic)\ne 0$
  3. If dissociation energy of $C{ H } _{ 4 }(g)$ is $1656kJ/mol$ and ${ C } _{ 2 }{ H } _{ 6 }(g)$ is $2812kJ/mol$, then value of $C-C$ bond energy will be $328kJ/mol$
  4. If ${ \Delta H } _{ f }({ H } _{ 2 }O,g)=-242kJ/mol; { \Delta H } _{ vap }({ H } _{ 2 }O,l)=44kJ/mol$ then ${ \Delta } _{ f }{ H }^{ o }({{OH}^{-}},aq)$ will be $-142kJ/mol$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

${ \Delta } _{ f }{ H }^{ o }(S,monolithic)\ne 0$ as it is not elemental form of S.
$\Delta E=0$ for combustion of ${ C } _{ 2 }{ H } _{ 6 }(g)$ in a sealed rigid adiabatic container as in adiabatic process energy exchange is zero.
For ${ C } _{ 2 }{ H } _{ 6 }(g)$,
$BE _{C-C} = 2812 - 6\times BE _{C-H} = 2812-6\times 1656/4 = 328$kJ/mol

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

Bond energies can be obtained by using the following relation:
$\Delta H$(reaction) $=\sum$ Bond energy of bonds, broken in the reactants  $- \sum$ Bond energy of bonds, formed in the products

Bond energy depends on three factors:
a. greater is the bond length, lesser is the bond energy
b. bond energy increases with the bond multiplicity
c. bond energy increases with the electronegativity difference between the bonding atoms.

Arrange $N-H$, $O-H$ and $F-H$ bonds in the decreasing order of bond energy:

  1. $F-H > O-H > N-H$
  2. $N-H > O-H > F-H$
  3. $O-H > N-H > F-H$
  4. $F-H > N-H > O-H$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Fluorine is more electron-negative than oxygen and oxygen is more electro-negative than nitrogen.

Hence, bond energy between $F-H$ is greater than $O-H$ which is greater than $N-H$.